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Pharmaceutical Statistics
Lecture 8
Distributions of the Difference
Between Two Sample Means
Δ
Distributions of the Difference Between Two Sample
Means
Comparison between two population is widely used and thus we need to learn how to
compare two sample means from two different population
Distributions of the Difference Between Two Sample
Means

Sample Mean
Distribution for
Population 1
X

Sample Mean
Distribution for
Population 2
X1
X2
X
1
X
2
 [X2 1   2 ]
1 X 2
X2
 [(2 /n )  (2 /n )]
1
Distribution of the difference
between two sample means
(population 1&2)
1
2
2
 SE()
X
1 X 2
 X 1 X 2
If we know the mean and the standard deviation of the new probability
distribution, we can use the SND approach to find out the probability
that the differences in the means will be less/more/between any specific
values (as we did before!!)
CASE OF B-A
CASE OF A-B

XA

XA
z
XA
X B
X B
     A  B

 [( 2 / n)  ( 2 / n)]

XA

X B
XB
A
A
B
B
XB
XB
(X A  X B )  X
X
A
X
A
X B
B
z
XB
X A
X A
     B  A
XB
XA
 [( 2 / n)  ( 2 / n)]

X A
B
B
A
A
(X B  X A )  X
X
B
X
B
A
X A
Note: this approach is valid for normally distributed populations even though they have
different known variances and/or different sample sizes
Note: For non-normally distributed populations, we need to draw large sample and thus
to apply CLT to assume population normality and finally to use this approach
Example I (Textbook, Chp 5, P143)
In two population: Population I has experienced some condition that
associated with mental retardation. The second population has not
experienced these conditions. The distribution of intelligence scores in
each of the two populations (1 and 2) is believed to be normally
distributed and equal for both with standard deviation of 20. A sample
of 15 individual from each population were withdrawn, compute the
probability of the difference between two means to be equal or larger
than 13?
X
1 X 2
X  X
1
2
First .case : x1  x2 13
 X 1  X2  1  

0
2
[(12 /n1)  (22 / n2 )]  SE()
 [(202 /15)  (202 /15)]
 7.3
z13 
(x1  x2 )  X  X
1
X
13
  1.78
7.3
2
1 X 2
Second.case : x2  x1 13....  x1  x2  13
z 1 3 
(x1  x2 )  X  X
1
X
1 X 2
2

13
 1.78
7.3
Distribution of the difference
between two sample means
(population 1&2)
X
1 X 2
 7.3
X X 0
1
2
SND
 1
AUC=0.0375
AUC=0.0375
Z=-1.78
z  0
Z=+1.78
From the textbook. P.143:
“The probability of obtaining a difference between sample means as or larger than 13 is
0.0375”…..
Do you agree?? I do not:
The probability should be 0.0375*2=0.075 since area to the right of z=+1.78 also satisfy the
condition that the difference is larger than 13!! The textbook took only case II in the
previous slide!
Example II
Population A: μA =45 min, σA=15 min
Population B: μB =43 min, σB=20 min
If we select 35 variables from pop A (sample A) and 40 variables from pop B
(sample B), what is the probability that the means for samples A&B will differ by 5
minutes or more?
A  45,A  15,nA  35
B  43,B  20, nB  40

 A  B 2
X X
A
X
A
A  45,A  15,nA 35
B  43,B  20,nB  40

 A  B  2
X A X B
B
X B
 [(A2 / nA)  ( B2 / nB )] 4.0

 [(A2 /n A )  (2B /n B )]  4.0
X A X B
firstcase :
sec ondcase:
X A  X B 5
X B  X A  5,then : X A  X B  5
5 2
 0.75
4.0
P(z  0.75)  1  P(z  0.75)
z5 
z5 
5 2
 1.75
4.0
P(z  1.75)  0.04
 0.23
Overall probability=0.23+0.04=0.27

Distribution of the difference
between two sample means
(populationA&B)
-5
X A X B

2
2
X A X B
4
XA  XB
5
SND
 1
AUC=0.23
AUC=0.04
-1.75
0
+0.75
z
X A X B
Overall probability=AUCright+AUCleft=0.23+0.04=0.27
Example III
Population A: μA =45 min, σA=15 min
Population B: μB =30 min, σB=20 min
If we select 35 variables from pop A (sample A) and 40 variables from pop B
(sample B), what is the probability that the of means for samples A&B will differ by
20 minutes or more?
A  45,A  15,nA  35
B  30,B  20,nB  40

 A  B 15
X X
A
X
B
A X B
 [(A2 / nA)  ( B2 / nB)]  4.05
A  45, A  15,nA  35
B  30, B  20,nB 40

 A  B 15
X A X B
X
A X B
 [( A2 / nA)  ( B2 / nB)]  4.05
firstcase :
sec ondcase:
X A  X B  20
X B  X A  20,then : X A  X B  20
20 15
1.23
4.05
P(z 1.23)  0.11
z 20 
20 15
 8.75
4.05
P(z  8.75)  zero
z20 
X
Distribution of the difference
between two sample means
(populationA&B)
-20
A
X
X B
A
X B
15
4
XA  XB
15 20
SND
The probability that the mean
of sample B>the mean of
sample A by 20 or more is
really low!!
 1
AUC=0.00
AUC=0.11
-8.75
0
+1.23
z
X A X B
Overall probability=AUCright+AUCleft=0.11+0.00=0.11
Example IV
The capsule weight for two hard-gelatin capsule batches A&B are normally distributed
with the following parameters:
- Batch A: μ=250 mg; σ=25 mg
- Batch B: μ=300 mg; σ=35 mg
If we withdraw a random sample from A (30 capsules) and a random sample from B
(40 capsules), what is the probability that the mean of sample B is larger than the
mean of sample A by at least 55 mg?
A  250, A  25,nA  30
B  300, B  35,nB  40
X X  B  A  50
B

X B X A
A

[(
2
B
/ n)
/ n)]
 7.17
B  (
A
2
A

X B X A
 50
X B  X A 55
55  50
 0.65
z 55 
7.71
AUC=0.26
Z=+0.65
Example IV (another way to solve it)
The capsule weight for two hard-gelatin capsule batches A&B are normally
distributed with the following parameters:
- Batch A: μ=250 mg; σ=25 mg
- Batch B: μ=300 mg; σ=35 mg
If we draw sample from A (30 capsules) and sample from B(40 capsules), what
is the probability that the mean of sample B is larger than the mean of
sample A by at least 55 mg?
A  250,A  25,nA  30
B  300,B  35,nB  40
X X  A  B  50
A

XA
B
X B
X
 [( 2 / n)  ( 2 / n)]  7.17
A
A
B
B
A X B
 50
X B  X A 55
X A  X B  55
55  (50)
 0.65
z55 
7.71
UC=0.26
Z=-0.65
Example IV (summary)
CASE OF B-A
CASE OF A-B
A  250, A  25,nA  30
B  300, B  35,nB  40
X X  A  B  50
A
X
A
B
X B
 [( / nA )  ( /n B )]  7.17
2
A
2
B
X B  X A  55
A
X B
X
 0.65
A
X B

A
 [( B2 / n B )  ( 2A / n A )]  7.17
X B X A
zX
 z55 
(X A  X B )  X
B
X B  X A  55
X A  X B  55
zX
A  250, A  25,nA  30
B  300, B  3 5,nB  40
X X  B  A  50
A XB
55  (50)

7.71
B
X A
 z 55 
(X B  X A )  X

 0.65
X B X A
B
X A

55  50
7.71
Pharmaceutical Statistics
Lecture 9
Estimating a Single Population Mean:
Point Estimate and Confidence Interval
15
Statistical Inferences
“Estimation”
From this lecture and on, we move from descriptive statistics to inferential
statistics.
In descriptive statistics, we simply summarize information available in the data
we are given. In inferential statistics, we draw conclusions about a population
based on a sample and a known or assumed sampling distribution. Implicit in
statistical inference is the assumption that the data were gathered as a
random sample from a population.
Statistical
Inferences
Estimation
Hypothesis
Testing
Prediction
16
The Process of Sample Infrence
{ESTIMATION}
17
Estimation of a Population Mean: Point Estimate and
Confidence Interval
• If we wish to estimate the mean of some normally distributed
population (μ), we would draw a random sample of size n from the
population and compute Xwhich can be used as a point estimate of
μ.
• Although the sample mean is a good estimator of the population
mean, we know that random sampling inherently involves chance
and the sample mean can not be expected to be equal to the
population mean.
• It is more meaningful to estimate μ by an interval that
communicates information regarding the probable magnitude of μ.
18
Estimation of a Population Mean: Point Estimate and
Confidence Interval
Sample taken from the
parent population
Sample mean:
The point estimate
Original variable distribution (X)
Sample mean
distribution

_
x
Sample Interval:
The Confidence Interval
19
Remember
Original variable distribution (X)
P(_  2 _  X  _  2_ ) 0.95
x
Sample mean
distribution
2_
2_
x
x
x
x
x
 
_
Constructing intervals about
every possible value of
computed mean from all
possible samples of size n
from the population of
interest. The width of each
interval is defined as 2 
x
X1
X2
X3
X4

x
X5
X6
In this case, we can say that
we are 95% confident that
the pop mean will fall in
randomly selected sample
intervals wit2hthe widthof

x
X7
X8

x  2

x
20
First Case:
Estimation of Population Mean if population variance is
known
n,x,
2
Sample mean +
size + population
variance/standard
deviation: known



x 2
x
x
Calculate mean
sampling
distribution
standard deviation
Use the C.I.
formula to
estimate the
population mean
Suppose a researcher, interested in obtaining an estimate of the average level of some
enzyme in a certain human population, takes a sample of 10 individuals, determines the level
of the enzyme in each and computes a sample mean of 22.
Suppose further it is known that the variable of interest is approximately normally
distributed with a variance of 45. Estimate the mean of the level of this enzyme (μ) in human
population?

x  22,2  45,n  10

x 2
x

x 2
2
n
45
10
22  2(2.1213)
22  2
17.76 26.24
Sample mean=22,
population
variance=45,
sample size=10
x±2σx
σx=2.1213
22±4.24
C.I: (17.76-26.24)
21
General Formula
x z(1 / 2)
z(1 /2)
x
•
X
is the point estimate of μ.
α/2
Cof. Level
1-α
α/2
• The “z(1-α/2)” is the valueof z to the left of which lies 1-α/2
and to the right of which lies α/2 of the AUC. This value called
the reliability coefficient
• (1-α) is called the confidence level
• x is the standard error or the standard deviation of the
sampling distribution of the sample mean.

C.I. : Estimator ∓ (Reliability Coefficient) x (Standard Error)
22
Common Levels of Confidence!!
z(1/2 )
α/2
Cof. Level
1-α
α/2
α
Conf. Level
1-α
0.1
90% (0.9)
0.05
95% (0.95)
0.01
99% (0.99)
z(1 / 2)
z0.95 1.645
z0.975 1.96
z0.995  2.576
 
90%CIof   X 1.645  n

95%CIof   X  1.96 
Any z can be used (from table)
n

 
99%CIof   X  2.576  n
In repeated sampling, approximately 90, 95, 99% of the intervals constructed with the
width (1.645SE, 1.96 SE, or 2.576SE respectively) will include the population mean
OR we are 90, 95, 99% confident that the population mean is enclosed within the
calculated confidence interval with the width (1.645SE, 1.96 SE, or 2.576SE
respectively).
23
Example
A physical therapist wished to estimate, with 99% confidence, the mean maximal
strength of a particular muscle in a certain group of individuals. Assuming that
strength scores are approximately normally distributed with a variance of 144. a
sample of 15 subjects who participated in the study yielded a mean of 84.3.
•
•
The z value corresponding to a confidence level of 0.99 is found to be 2.58
(reliability coefficient is ???).
The 99% CI for μ is
 
99%CIof   X  2.58  n
 84.3  2.58(12/15)
 84.3  8
99%CIof   76.3  92.3
24
Pharmaceutical Statistics
Lecture 10
Estimating a Single Population Mean:
The t-distribution
The t Distribution
• If the population standard deviation σ is known, we have learned last
lecture how to estimate a population mean based on sample mean.
• However, usually the population standard deviation is unknown as well as
the population mean (what do we need to do in this case?).
• We may use the sample standard deviation to replace σ. Here the zstatistics will convert to t-statistics :
Note:
-When we have small
samples, it becomes
necessary for us to use the tdistribution in constructing
the confidence interval.
- When the sample size is
large (>30), our faith in s as
an approximation of σ is
usually substantial, and we
may be justified in using
standard normal distribution
theory to construct a
confidence interval for the
population mean.
Standard Normal Distribution

x 
z 
 / n
z-distribution if σ is known
Student’s t distribution

t 
x 
s / n
t-distribution if σ is unknown
[pop variance is unknown+small n]
The t-Distribution: Properties
The variable t ranges from  to  
It has a mean of 0
It is symmetrical about the mean
In general it has a variance greater than 1, but the variance
approaches 1 as the sample size becomes larger.
5. Compared to the normal distribution, the t distribution is less
peaked in the center and has higher tails.
6. The t distribution approaches the normal distribution as n
increases.
7. Degree of Freedom=n-1
1.
2.
3.
4.
The t Distribution table
• The t distribution, like the SND, has been extensively tabulated.
• We must take both the cumulativ probability (AUC-∞tdf)and degrees of
freedom into account to find t-scores (critical values) when using the table
of the t distribution.
D.F=n-1=11
Note: differentiate between cum.
Probability and confidence level
Z- versus t-table
Z-table
Only one parameter: z-score
T-table
Two parameters: cumm.prop and
D.F
We calculate z-score and then we
find probability (AUC)
We decide on the cumm.prop
(from Conf. Level) and then we
find the t-score (reliability
coefficient)
Always entries less than the
When n approach ∞, t-table
corresponding entries for t-tables entries match the z-table entries
at same probability
For the z-curve: μ=0; σ=1
For the t-curve: μ=0; σ>1 and
approach 1 when n is large
The t Distribution

The general procedure for constructing confidence intervals
is not affected by us having to use the t distribution rather
than SND.
estimator ∓ (reliability coefficient) x (standard error)
X  z(1 / 2) *

n
s
X  t(1 /2) *
n
or
X  z(1 / 2) * SE
o
r
X  t (1 /2) * SE
Example
20 tablets were chosen randomly from a batch. Their weights in
mg were
300 321 306 321 310 322 315 325 316 323 316 325
317 325 319 327 320 331 320 336. Estimate the batch mean
weights with a confidence level of 95%?
- Sample mean and standard deviation t-distribution: σ is unknown+small n
x  319.75mg
n
s

 xi  x
i1

n 1
2
 8.2
- df is 19 so t*=2.093 (t-table)
95%CIof  
 
3 1 9 . 7 5  2.093 s n
=315.91-323.59 mg
Should I use z- or t-table?
HW5
Q1: We wish to estimate the average number of heartbeats per minute for a certain
population. The average number of heartbeats per minute for a sample of 49
subjects was found to be 90. Assume that these 49 patients constitute a random
sample, and that the population is normally distributed with a standard deviation
of 10.
1. Construct 90, 95, and 99% C. I for the population mean?
2. How do you interpretate these C. I(s)?
3. Which one do you prefer to use? Why?
4. How does sample size affect the width of the CI(s)? Consider n= 49 and then
490?
Q2. Use the t-distribution to find the reliability factor for a confidence interval based
on the following confidence coefficient and sample size:
a
b
c
d
Confidence level 0.95
0.99
0.9
0.95
Sample size
24
8
30
15
Cont..HW5
•
•
Q3. A sample of 16 girls had a mean weight of 71.5 and a standard deviation of 12
pounds respectively. Assuming normal distribution, find the 90, 95, 99 %
confidence intervals for μ?
Q4. A simple random sample of 16 subjects yielded the following values of urine
excreted arsenic (mg/day):
Subject Value
Subject Value
1
0.007
9
0.012
2
0.030
10
0.006
3
0.025
11
0.010
4
0.008
12
0.032
5
0.030
13
0.006
6
0.038
14
0.009
7
0.007
15
0.014
8
0.005
16
0.011
Construct a 95% confidence intervals in μg/day
for the population mean
Pharmaceutical Statistics
Lecture 11
Estimating The Difference between Two Population
Means
Estimating The Difference between Two Population Means
• In the previous lectures, we learned how to estimate the mean of single
population and how to construct C.I for the mean.
• In some cases we are interested in estimating the difference between
means of two populations.
• This estimation helps us in deciding whether or not it is likely that the
two populations means are equal.
• If the constructed confidence interval does include “zero” we can say that
the populations may be equal, and vise versa.
0
20
The C.I does
not include
zero, thus
population
means may
NOT be equal
The C.I does
include zero,
thus population
means may be
equal
0 2
Sample from pop 1
(n1,X1 ,s1 )
Sample from pop 2
(n2,X 2 , s2)
C.I for the sample mean differences
0
20
The C.I does
not include
zero, thus
population
means may
NOT be equal
The C.I does
include zero,
thus population
means may be
equal
0 2
A) When we withdraw two samples from two populations that
are normally distributed with known σ1 and σ2
• Normal distribution populations
• Population variances are known
• Small or large samples
• z-table
C.I. : Estimator ∓ (Reliability Coefficient) x (Standard Error)
C.I  (X 1  X 2 )  z(1 / 2)X X
1
X
1
X
2
2
 [(12 /n1)  (22 /n 2 )]
Example
•
A research tem is interested in the difference between two serum uric acid levels
in patients with or without Dawn’s syndrome. A sample of 12 individual with
Dawn’s syndromes yielded a mean of X1=4.5 mg/100 mL. A sample of 15 normal
individuals of the same age and sex were found to have a mean value of X 2 =3.4
mg/100 mL. Assume that the two populations are normally distributed. With
variances equal to 1 and 1.5 (Dawn’s and normal, respectively). Construct the 95 %
confidence interval for the Dawn's  Normal
(X 1  X 2 )  z(1 / 2)XX
1
2
(X 1  X 2 )  4.5  3.4 1.1
At 95% confidence level, the reliability coefficient from the z-table is 1.96
X
1
X
2
 [(12 /n1)  (22 / n2 )]  [(1/12)  (1.5/15)]  0.4282
C.I : (X 1  X 2 )  z(1 / 2)X
 1.1  1.96(0.4282)
 1.1  0.84
(0.261.94)
1
X
2
What do you conclude from this
confidence interval?? we can conclude that
the population means may be not equal with
a confidence level of 95% [zero is not
included in the C.I.
B) When we withdraw two samples from two populations that are
NOT normally distributed (σ1 and σ2 are unknown) and the sample
size is large
If we have no idea about the normality of the parent distributions and their parameters, we use
the CLT to justify our use of the z-table to find the reliability coefficient. We use the samples
standard deviations to calculate the standard error of the difference. This is only true for
large samples
Example: Two samples from two populations were withdrawn. The first sample (112 subjects)
from the first pop yielded a mean of 401.8 and a standard deviation of 226.4. The second
sample (75 subjects) from the second pop yielded a mean of 828.2 and a standard
deviation of 274.9. Construct the 99% C.I for the difference between population
means?
• We have no idea about the normality of the parent populations
•
Since the samples are large, we can apply the CLT and assume that
the distribution of the sample mean difference is normally
distributed
• We can use the z-table to find the reliability coefficient and the standard
deviation values of the samples to calculate the standard error of the
mean differences
B) When we withdraw two samples from two populations that are
NOT normally distributed (σ1 and σ2 are unknown) and the sample
size is large
C.I  (X 1  X 2 )  z(1 / 2)X
X
1
X
2
1
X
2
 Xs X  [(s12 /n1 )  (s22 /n2)]
1
2
(X 1  X 2 )  828.2226.4  426.4

X
1
X
s
2
X 1 X
Due to large n
2
2
 [(s12 /n1 )  (s/
[(274.9)2 /75  (226.42 ) /112]  38.2786
2 n2 )] 
C.I : (X 1  X 2 )  z(1 / 2)X
 42 6.4  2.58(38.2786)
 426.4  98.76
1
X
2
Due to CLT
 (32 7.6  52 5.2)
We have no idea about the normality of the parent populations, but because the samples size is
large, we applied the CLT and assumed that the distribution of the sample mean difference is
normally distributed. With this in mind, we used the z-table to find the reliability coefficient
(2.58). Also we used the standard deviation values for both samples to calculate the standard
error of the means difference (38.2786). The interval does not include zero, so we can
conclude that the population means may be not equal with a confidence level of 99%.
C) When we withdraw two samples from two populations that are
normally distributed (σ1 and σ2 are unknown) and the sample size is
small
• Parent pops: Normally distributed
• Variance for pops: Unknown
• Sample size: Small
We can not use the z-distribution in this case. We need to use the t-table
to find the reliability factor and we need to use standard deviations of
the two samples to find the standard error
Important note in this case:
The calculation of the standard error from s1,n1 and s2, n2 depend on the equality of
the parent populations variances
C.1) If they are
equal: we use tdistribution
C.2) If they are
not equal: we use
the t’-distribution
C.1) When we withdraw two samples from two populations that are normally
distributed (σ1 and σ2 are unknown and equal) and the sample size is small
•
•
In this case, we need to consider the samples variances+ to use t-table.
Since the sample variance is dependent on the sample size, we need to take this in
consideration for samples with different sizes (n) to calculate the pooled estimate
of the common variance
[This formula pools both sample
2
2
(n1 1)s1  (n 2 1)s2
s 
n1  2n 2
variances with their corresponding
weight that based on the sample
size]
2
p
•
NOTE: If the sample size for both independent samples is equal, we take directly
the arithmetic mean of the two samples variances (simple average) .
The standard error of the estimate will be:
s 2p s 2p

X 1 X
•
s
2
X 1 X 2

n1

n2
The 100(1-α)% confidence interval for the mean difference will be:
C.I : (X 1  X 2 )  t(1 / 2)X
 (X 1  X 2 )  t (1 /2)
s 2p
n1
p
2
1
2

X
s2
n
We use the t-table with D.F= n1+n2-2 to
find the reliability coefficient
Example
• Two independent samples were taken from normally distributed populations
A and B that have equal variance. Sample from A (n=13) resulted in a mean
value of 21 and standard deviation of 4.9. Sample from B (n=17) resulted in a
mean value of 12.1 and standard deviation of 5.6. Construct 95% confidence
interval for the difference between the means of population A and B.
Since A&B are normally distributed, their
variances are unknown and equal, and the
sample size is small we use the t-table and
pooled estimate of the common
variance
2
2
(n
1)s
(n
1)s
1
2
2
s2p  1
n1  n2  2

X1 X
s
2
X 1 X 2
 [(s /n1 )  (s/ n2)]
2
p
C.I  (X 1  X 2 )  t(1 / 2)X
2
p
1
X
2
n1  13,s1  4.9
n2  17,s2 5.6
.............................................................
s 
(13 1)4.9 2  (17 1)5.6 2
 28.21
13 17  2
28.21 28.21

 1.957
sX 1 X 2 
13
17
.............................................................
2
p
C.I  (X 1  X 2 )  t(1 / 2)
s 2p
n1

s 2p
n2
 (21 12.1)  2.045 * (1.957)
 8.9  4.0085
 (4.9 12.9) What does this mean??
C.2) When we withdraw two samples from two populations that are
normally distributed (σ1 and σ2 are unknown and NOT equal) and the
sample size is small
• We can not use the t-table to find the reliability factors for
D.F= n1+n2-2.
• Solution: Instead of finding the reliability factor, we need to
compute it taking into consideration the reliability factor for
each sample and the weight of each reliability factor.
How to compute the reliability factor?
t' 1 /2 
w1t1  w 2 t 2
w1  w
s12
w1 
n1
2
s22
w2 
n2
t1  t1 / 2 ....... for: n1 1
t2  t1 / 2 ....... for: n2 1
100(1-α)% Confidence interval formula for μ1-μ2
'
C.I  (X
X)  t(1
1
 /2)
2
s2  s22
1
n1 n 2
Example
• Two independent samples were taken from normally distributed
populations A and B. We do not have any valid reason to assume the
equality of the population variance for A&B. Sample from A (n=13)
resulted in a mean value of 4.5 and standard deviation of 0.3. Sample from
B (n=17) resulted in a mean value of 3.7 and standard deviation of 1.0.
Construct 95% confidence interval for the difference between the means
of population A and B.
0.32
 0.007
13
12
w2   0.06
17
t1  t1 / 2 ....... for : n1 1  2.1788
w1 
t2  t1 / 2 ....... for : n2 1  2.199
t' 1 /2
w t  w2t2
 11
w1  w 2
(0.007  2.1788)  (0.06  2.199)

0.007  0.006
 2.1262
C.I  (X 1  X 2 )  t
'
(1 /2)
 (4.5  3.7)  2.1262
s12 s22

n1 n 2
0.3
2
13

12
17
 0.8  2.1262(0.256)
 (0.25 1.34)
What do you conclude from the
confidence interval above??
Should I use z, t, or t’ table?
Summary: Estimating The Difference between Two Population Means